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Math Help - Chain rule question...

  1. #1
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    Chain rule question...

    Given the function:
    s(t)=\sin2\pi*t

    My book says: s'(t)=2\pi*\cos2\pi*t

    I'm pretty sure this is a utilization of the chain rule, but I don't really understand it, can someone clear it up for me?

    (ps, sorry I could not get it to show up any clearer...)
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  2. #2
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    chain rule is g'(h(x)) h'(x) . which means..

    g(x)=sin(x) [x being sin(2Pi)]
    g'(x)=cos(x)
    h(x)=sin(2Pi) and
    h'(x)=0 .

    put it all into the function and you get..

    cos(sin(2Pi)) (0)

    which ends up giving you a 0 anyways. are you sure thats what the book is saying? lol i mean it sounds dumb to ask but this is the way i did it.. idk and i tried many times on the calculator and i keep getting zero because the derivative of t is 1.. and anything that you multiply by zero stays zero... you know??? any more info you have?
    Last edited by calcgeek_17; March 30th 2009 at 04:53 PM. Reason: my bad. der of sin is not (-) cos just cos. =]
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  3. #3
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    Quote Originally Posted by rust1477 View Post
    Given the function:
    s(t)=\sin2\pi*t

    My book says: s'(t)=2\pi*\cos2\pi*t

    I'm pretty sure this is a utilization of the chain rule, but I don't really understand it, can someone clear it up for me?

    (ps, sorry I could not get it to show up any clearer...)
    Is s(t) = sin(2pi times t), or

    sin(2pi) times t?

    makes it easier for me to help you hehe
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  4. #4
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    yea! lol.. im pretty sure thats it.. that makes more sense. hehe how come i didnt see that?? good job.
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  5. #5
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    Quote Originally Posted by rust1477 View Post
    Given the function:
    s(t)=\sin2\pi*t

    My book says: s'(t)=2\pi*\cos2\pi*t

    I'm pretty sure this is a utilization of the chain rule, but I don't really understand it, can someone clear it up for me?

    (ps, sorry I could not get it to show up any clearer...)
    Clearly the function is s(t) = \sin (2 \pi t). Please post your equations so that they are less ambiguous.

    You're expected to know the standard form: If y = \sin (kt) then \frac{dy}{dt} = k \cos (kt).
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